Computing the limit $\lim_{x\to \infty } \frac{x^2(1+\sin^2(x))}{(x+\sin(x))^2} $ Can someone please help me with this limit? Assuming that in class we have not used derivatives, so I can't use it.
$$\lim_{x\to \infty } \frac{x^2(1+\sin^2(x))}{(x+\sin(x))^2} $$ 
I have that is the same as
$$\lim_{x\to \infty } \frac{1+\sin^2(x)}{1+ \frac{2\sin(x)}{x} + \frac{\sin^2(x)}{x^2}} $$ 
And I do not know how to proceed.
 A: A hint: when $x$ is large, then $\sin(x) / x$ is small in absolute value -- say, $|\sin(x) / x| < 0.00001$, although you could make it as small as you wish. Thus, the corresponding terms become insignificant, and the entire fraction is very close to $\frac{1 + \sin^2 (x)}{1}$. What does that fraction do as $x$ becomes large?
A: *

*By putting $x:=n\pi$, for $n=1,2,3,\cdots,$ one gets $$ \lim_{x\to
   \infty } \frac{x^2(1+\sin^2(x))}{(x+\sin(x))^2}=\lim_{x\to \infty }
   \frac{x^2(1+0)}{(x+0)^2}=1. $$

*By putting $x:=(2n+1)\dfrac\pi2$, for $n=1,2,3,\cdots,$ one gets $$  
   \lim_{x\to \infty } \frac{x^2(1+\sin^2(x))}{(x+\sin(x))^2}=\lim_{x\to
   \infty } \frac{x^2\cdot 2}{(x\pm1)^2}=2, $$
Thus the limit does not exist.
A: Let $f$ be the function we want to compute the limit. If you take $x_n = n\pi$ since $\sin (x_n) = 0$, $$\lim_{n\to\infty} f(x_n) = 1$$ and if $y_n = \frac\pi2 + 2n\pi$, $\sin (y_n) = 1$ $$\lim_{n\to\infty} f(y_n) = 2$$ Which proves that $f$ does not have a limit when $x\to\infty$.
